pysipp - Trying to use it with existing sipp conf file

Question

Background

I have an existing sipp conf file that I launch like so:

sipp mysipdomain.net -sf ./testcall.conf -m 1 -s 12345 -i 10.1.1.1:5060

This runs just fine. It simulates a call in our test labs. But now I need to expand this test to make it a part of a larger test script where not only do I launch the sipp test, but I prove (via sip trace) that it's hitting the right boxes.

I decided to wrap this sipp call in python. I just found https://github.com/SIPp/pysipp and am trying to see if I can write this entire test in python. To start, i tried to run the same sipp test using pysipp.

Problem / Question

I'm currently getting an error that says:

lab2:/tmp/jj/sipp_tests# python mvv_numeric.py
No handlers could be found for logger "pysipp"
Traceback (most recent call last):
  File "mvv_numeric.py", line 6, in <module>
    uac()
  File "/usr/lib/python2.7/site-packages/pysipp-0.1.alpha-py2.7.egg/pysipp/agent.py", line 71, in __call__
    raise_exc=raise_exc, **kwargs
  File "/usr/lib/python2.7/site-packages/pluggy-0.3.1-py2.7.egg/pluggy.py", line 724, in __call__
    return self._hookexec(self, self._nonwrappers + self._wrappers, kwargs)
  File "/usr/lib/python2.7/site-packages/pluggy-0.3.1-py2.7.egg/pluggy.py", line 338, in _hookexec
    return self._inner_hookexec(hook, methods, kwargs)
  File "/usr/lib/python2.7/site-packages/pluggy-0.3.1-py2.7.egg/pluggy.py", line 333, in <lambda>
    _MultiCall(methods, kwargs, hook.spec_opts).execute()
  File "/usr/lib/python2.7/site-packages/pluggy-0.3.1-py2.7.egg/pluggy.py", line 596, in execute
    res = hook_impl.function(*args)
  File "/usr/lib/python2.7/site-packages/pysipp-0.1.alpha-py2.7.egg/pysipp/__init__.py", line 250, in pysipp_run_protocol
    finalize(cmds2procs, raise_exc=raise_exc)
  File "/usr/lib/python2.7/site-packages/pysipp-0.1.alpha-py2.7.egg/pysipp/__init__.py", line 228, in finalize
    raise SIPpFailure(msg)
pysipp.SIPpFailure: Some agents failed
'uac' with exit code 255 -> Command or syntax error: check stderr output

Code

Here's what the py script looks like:

  1 import pysipp
  2 uac = pysipp.client(destaddr=('mysipdomain.net', 5060))
  3 uac.uri_username = '12345'
  4 uac.auth_password = ''
  5 uac.scen_file = './numeric.xml'
  6 uac()

And the original sipp "testcall.conf" has been renamed to "numeric.xml" and looks like this: (I'm only including the first part because it's quite long. if you need to see something specific, please let me know and I will add to this post)

<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE scenario SYSTEM "sipp.dtd">
<scenario name="UAC with Media">
<send retrans="10000">
<![CDATA[

      INVITE sip:[service]@[remote_ip]:[remote_port] SIP/2.0
      Via: SIP/2.0/[transport] [local_ip]:[local_port];branch=[branch]
      From: sipp <sip:sipp@[local_ip]:[local_port]>;tag=[pid]SIPpTag00[call_number]
      To: [service] <sip:[service]@[remote_ip]:[remote_port]>
      Call-id: [call_id]
      CSeq: 1 INVITE
      Contact: <sip:sipp@[local_ip]:[local_port]>
      Allow: INVITE, ACK, BYE, CANCEL, OPTIONS, INFO, MESSAGE, SUBSCRIBE, NOTIFY, PRACK, UPDATE, REFER
      User-Agent: PolycomVVX-VVX_300-UA/5.5.2.8571
      Accept-Language: en
      Supported: replaces,100rel
      Allow-Events: conference,talk,hold
      Max-Forwards: 70
      Content-Type: application/sdp
      Content-Length: [len]

I'm sure it's something simple I've overlooked. Any pointers would be appreciated.

EDIT:

I added debug level logging and reran the python script. In the logs I can now see what pysipp is actually attempting:

2018-01-31 14:40:32,715 MainThread [DEBUG] pysipp launch.py:63 : launching cmd:
"'/usr/bin/sipp' 'mysipdomain.net':'5060' -s '12345' -sn 'uac' -sf 'numeric.xml' -recv_timeout '5000' -r '1' -l '1' -m '1' -log_file '/tmp/uac_log_file' -screen_file '/tmp/uac_screen_file' -trace_logs -trace_screen "

So comparing that with the original command line I use to run sipp, I see the extra "-sn 'uac'".

Going to see about either getting my SIPP script to work with that tag or ... google to see if I can find other similar posts. In the meantime, if you see my mistake, i'm all ears.

next time an issue on the Github project would be useful as I'm not sure the SIPp team hangs on SO that much unfortunately. — goodboy, May 18 '18 at 04:17

score 2 · Answer 1 · edited Jan 06 '20 at 13:02

The problem here (as you noticed) is likely that pysipp.client() sets the -sn uac flag and sipp fails having both -sn and -sf.

To see the actual error you can enable logging before running the client:

import pysipp

pysipp.utils.log_to_stderr("DEBUG")

uac = pysipp.client(destaddr=('mysipdomain.net', 5060))
uac.uri_username = '12345'
uac.auth_password = ''
uac.scen_file = './numeric.xml'
uac()

The hack is to simply do uac.scen_name = None but the proper way to do this is to either use pysipp.scenario() (docs here) and rename your numeric.xml to have uac in the file name (i.e. uac_numeric.xml) or use instead pysipp.ua(scen_file=<path/to/numeric.xml>).

To understand the problem the client is currently applying a default scenario name argument when really the user should be able to override that (though in that case there'll be no guarantee that the user actually is sending client traffic which renders the name client kind of pointless).

pysipp - Trying to use it with existing sipp conf file

1 Answers1